With each square matrix corresponds just one number. This number is called the determinant of the matrix. The determinant of a matrix A is denoted det(A) or |A|. Now we'll define this correspondence.
|A| = sum sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} . ... . a_{n,t(n)} t in P_{S}Note that each term of |A| uses each row and each column only once.
Example1 : We want to calculate the determinant of a 2x2 matrix A.
Now n = 2 and there are only two permutations of S = (1,2).
t(1,2) = (1,2) with sgn(t) = +1
t'(1,2) = (2,1) with sign(t') = -1
We have only two terms +1.a_{1,1} . a_{2,2} and -1.a_{1,2} . a_{2,1}
Thus the determinant of A is a_{1,1} . a_{2,2} - a_{1,2} . a_{2,1}
We don't forget the rule :
|a b| |c d| = ad - cb
Example2 : We want to calculate the determinant of a 3x3 matrix A.
Now n = 3 and there are only 6 permutations of S = (1,2,3).
These 6 permutations transform (1, 2, 3) in:
(1, 2, 3) (2, 3, 1) (3, 1, 2) (even permutations) (3, 2, 1) (1, 3, 2) (2, 1, 3) (odd permutations)Now we have six terms to add
a_{1,1} . a_{2,2} . a_{3,3} + a_{1,2} . a_{2,3} . a_{3,1} + a_{1,3} . a_{2,1} . a_{3,2} -a_{1,3} . a_{2,2} . a_{3,1} - a_{1,1} . a_{2,3} . a_{3,2} - a_{1,2} . a_{2,1} . a_{3,3}We don't forget the rule :
|a b c| |d e f| |g h i| = aei + bfg + cdh - ceg - afh - bdiThe last rule is known as the Sarrus rule for 3 x 3 determinants.
To calculate larger determinants there are a lot of other methods involving various properties of determinants.
Now consider A_{i,j} . a_{i,j}. This term uses the i-th row and the j-th column in the factor
a_{i,j}. So the cofactor A_{i,j} is independent of the elements of the i-th row and the
elements of the j-th column.
The cofactor A_{i,j} contains only elements from the matrix obtained from A by crossing
out the i-th row and the j-th column.
Remark: If we write |A| = A_{i,1} . a_{i,1} + A_{i,2} . a_{i,2} + A_{i,3} . a_{i,3} + ... A_{i,n} . a_{i,n}, we say that the determinant is calculated emanating from the i-th row.
Example :
|a b c| |d e f| |g h i| = aei + bfg + cdh - ceg - afh - bdiChoose for instance row 2.
If we write |A| = (ch-bi)d + (ai-cg)e + (bg-ah)f , we say that the determinant is calculated emanating from the second row.
Similarly, we can start with a fixed column and then write |A| as a linear polynomial in a_{1, j}, a_{2,j}, a_{3,j}, ... a_{n,j}. Then one finds the same cofactors. So a_{i,j} has a unique cofactor A_{i,j}.
|B| = sum sgn(t). b_{1,t(1)} . b_{2,t(2)} . b_{3,t(3)} ... b_{n,t(n)} t in P_{S} but b_{i,j} = a_{j,i} = sum sgn(t). a_{t(1),1} . a_{t(2),2} . a_{t(3),3} ... a_{t(n),n}. t in P_{S}Consider the last sum.
Then
|B| = sum sgn(t') . a_{1,t'(1)} . a_{2,t'(2)} . a_{3,t'(3)} ... a_{n,t'(n)} t' inverse of t Because the set of all permutations of S is the same set of all inverse permutations. So we can write |B| = sum sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... a_{n,t(n)} t in P_{S} So, |B| = |A|.
Sgn(t') = -1 and for each permutation t we have sign(t't) = -sign(t).
Say B is obtained by interchanging the column i and j of A.
For each k we have b_{k,i} = a_{k,j} and b_{k,j} = a_{k,i} or even for each k and each l we have
b_{k,l} = a_{k,t'(l)}We investigate |B|.
|B| = sum sgn(t). b_{1,t(1)} . b_{2,t(2)} . b_{3,t(3)} ... b_{n,t(n)} t in P_{S} but b_{k,l} = a_{k,t'(l)} |B| = sum sgn(t). a_{1,t't(1)} . a_{2,t't(2)} . a_{3,t't(3)} ... a_{n,t't(n)} t in P_{S} |B| = sum -sgn(t't). a_{1,t't(1)} . a_{2,t't(2)} . a_{3,t't(3)} ... a_{n,t't(n)} t in P_{S}Since the set of permutations of (1 ... n) is a group, the set of all permutations t and the set of all permutations t" = t't is the same set. Therefore
|B| = sum -sgn(t"). a_{1,t"(1)} . a_{2,t"(2)} . a_{3,t"(3)} ... a_{n,t"(n)} t" in P_{S} |B| = sum -sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... a_{n,t(n)} t in P_{S} |B| = -|A|Conclusion :
= sum sgn(t) . b_{1,t(1)} . b_{2,t(2)} . b_{3,t(3)} ... b_{i,t(i)} ... b_{n,t(n)} t in P_{S} = sum sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... r.a_{i,t(i)} ... a_{n,t(n)} t in P_{S} = sum r sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... a_{i,t(i)} ... a_{n,t(n)} t in P_{S} = r.( sum sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... a_{i,t(i)} ... a_{n,t(n)}) t in P_{S} = r.|A|Conclusion :
|A| = sum sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... a_{i,t(i)} ... a_{n,t(n)} t in P_{S} |B| = sum sgn(t) . b_{1,t(1)} . b_{2,t(2)} . b_{3,t(3)} ... b_{i,t(i)} ... b_{n,t(n)} t in P_{S} = sum sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... b_{i,t(i)} ... a_{n,t(n)} t in P_{S} So |A| + |B| = sum sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... (a_{i,t(i)} + b_{i,t(i)}) ... a_{n,t(n)} t in P_{S} = determinant of the matrix formed by adding the i-th row from A and B, and taking the other elements from A or from B.The same rule holds for columns
|a b c| |a b' c| |a b+b' c| |d e f|+|d e' f| = |d e+e' f| |g h i| |g h' i| |g h+h' i|
| a_{1,1} a_{1,2} .... | a_{2,1} a_{2,2} .... | |A| = | | a_{i,1} a_{i,2} .... | | | a_{j,1} a_{j,2} .... |Let B be the matrix formed by replacing in A the i-th row by the j-th row, leaving the jth row unchanged.
| a_{1,1} a_{1,2} .... | a_{2,1} a_{2,2} .... | |B| = | | a_{j,1} a_{j,2} .... | | | a_{j,1} a_{j,2} .... |Since B has two equal rows |B| = 0.
| a_{1,1} a_{1,2} .... | a_{2,1} a_{2,2} .... | |C| = | |r.a_{j,1} r.a_{j,2} .... | | | a_{j,1} a_{j,2} .... |Then |C| = r.|B|=0
| a_{1,1} a_{1,2} .... | a_{2,1} a_{2,2} .... | |A|+|C| = | |a_{j,1}+r.a_{j,1} a_{j,2}+r.a_{j,2} .... | | | a_{j,1} a_{j,2} .... |But since |C| = 0 we have |A|+|C| = |A| + 0 = |A|. So, the last determinant is equal to the first one.
Therefore, a determinant does not change if we add a multiple of a row to another row.
The same rule holds for columns
Ex.
|a b c| |a+rd b+re c+rf| |d e f| = | d e f | |g h i| | g h i |
|A| = sum sgn(t) . a_{1,t(1)} . a_{2,t(2)} . a_{3,t(3)} ... a_{n,t(n)}. t in P_{S}A_{1,1} is the coefficient of a_{1,1} in this sum.
Now, instead of taking the sum for t in P_{S} we only take the sum for t in the set
{t in P_{S} | t(1) = 1} = permutations of S' =(2 ... n).
This sum then gives exactly A_{1,1} . a_{1,1}.
A_{1,1} . a_{1,1} = sum sgn(t) . a_{1,1} . a_{2,t(2)} . a_{3,t(3)} ... a_{n,t(n)} t in P_{S}' = a_{1,1} ( sum sgn(t) a_{2,t(2)} . a_{3,t(3)} ... a_{n,t(n)} ) t in P_{S}' Thus A_{1,1} = sum sgn(t) a_{2,t(2)} . a_{3,t(3)} ... a_{n,t(n)} t in P_{S}'This last sum is, by the definition of a determinant, the determinant of the sub-matrix of A obtained from A by crossing out the first row and the first column.
Conclusion:
A_{1,1} = the determinant of the sub-matrix of A obtained from A by crossing out the first row and the first column.
Then we interchange in succession column j and j-1; j-1 and j-2; ... until e is on the first column and on the first row. This demands j-1 steps.
During this process the determinant of the matrix changes i+j-2 times sign.
Now the cofactor of e is the determinant of the sub-matrix of obtained from by crossing out the first row and the first column.
Now we return to the original matrix.
The value of A_{i,j} = (-1)^{i+j-2}.(the determinant of the sub-matrix of A, obtained from A by crossing out the i-th row and the j-th column).
Or stated simpler:
The value of A_{i, j} = (-1)^{i+j}.(the determinant of the sub-matrix of A, obtained from A by crossing out the i-th row and the j-th column).
Let I be the (k+1) x (k+1) identity matrix and we calculate this matrix emanating from the first row.
|A| = A_{1,1} . a_{1,1} + A_{1,2} . a_{1,2} + A_{1,3} . a_{1,3} + ... A_{1,n} . a_{1,n}.
|A| = A_{1,1}.1 + A_{1,2}.0 + A_{1,3}.0 + ... A_{1,n}.0.
|A| = A_{1,1}
Now, the cofactor A_{1,1} is the determinant of the k x k identity matrix, and this determinant is 1.
|a b| |c d| = ad - cb
|a b c| |d e f| |g h i| = aei + bfg + cdh - ceg - afh - bdi
Example
Take the 3 x 3 matrix A =
[4 5 7] [1 2 3] [2 5 6]We calculate the cofactor corresponding with the element a_{1,3} = 7.
|A| = A_{i,1} . a_{i,1} + A_{i,2} . a_{i,2} + A_{i,3} . a_{i,3} + ... A_{i,n} . a_{i,n}
A_{i, j} is the cofactor of a_{i,j}.
We say we have unfold the determinant following row i.
Example
Take the 3 x 3 matrix A =
[4 5 7] [1 2 3] [2 5 6]We unfold the determinant following row 1.
We unfold the determinant following column 3.
The three cofactors are 1, -10, 3.
|A| = 7.1 - 3.10 + 6.3 = -5
|a b c| |a b' c| |a b+b' c| |d e f|+|d e' f| = |d e+e' f| |g h i| |g h' i| |g h+h' i|
R3 means row 3 K2 means column 2 R2 - R3 means: replace R2 by R2 - R3 (the value of the determinant does not change) K2 + 2K3 means: replace K2 by K2 + 2.K3 (the value of the determinant does not change)
| x 2m 1 | | 3 1 1 | = | x m 1 | R1 - R3 (create zeros) | 0 m 0 | | 3 1 1 | = | x m 1 | unfold the determinant following row 1 m.( -(3-x)) = m(x-3) -------------------------------- | 1 2 3 | | 4 5 6 | = | 7 8 9 | R1 - R2 ; R2 - R3 | -3 -3 -3| | -3 -3 -3| = 0 | 7 8 9 | --------------------------------- | 1 1+m -1 | | 3 3+m -3 | = | 5 m -1 | K1 + K3 (create zeros) | 0 1+m -1 | | 0 3+m -3 | = | 4 m -1 | unfold the determinant following column 1 0 + 0 + 4.((1+m).(-3) + (3+m)) = -8m ------------------------------- | m^{2} + m m m^{3}| | a b c | = | d e f | factor m in the first row | m + 1 1 m^{2}| m.| a b c | = | d e f |
|ra rb rc| |a b c| |d e f| =r.|d e f| |g h i| |g h i|We reformulate this property :
Example
| a a^{2} a^{3} | | b b^{2} b^{3} | = | c c^{2} c^{3} | factor a in row 1 factor b in row 2 factor c in row 3 | 1 a a^{2} | a.b.c.| 1 b b^{2} | = | 1 c c^{2} | R1 - R2 ; R2 - R3 (create zeros) | 0 a-b a^{2}-b^{2} | a.b.c.| 0 b-c b^{2}-c^{2} | = | 1 c c^{2} | factor in row 1 factor in row 2 | 0 1 a+b | a.b.c.(a-b).(b-c).| 0 1 b+c | = | 1 c c^{2} | unfold following the first column a.b.c.(a-b).(b-c).(c-a)